Let `vecn` be a unit vector perpendicular to the plane
containing the point whose position vectors are
`veca, vecb and vecc` and, if
`abs([(vecr -veca)(vecb-veca)vecn])=lambda abs(vecrxx(vecb-veca)-vecaxxvecb)` then `lambda` is
equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to analyze the given condition and derive the value of \(\lambda\).

### Step 1: Write down the given condition
We start with the condition provided in the question:
\[ |\vec{r} - \vec{a}| \cdot |\vec{b} - \vec{a}| \cdot \vec{n} = \lambda |\vec{r} \times (\vec{b} - \vec{a}) - \vec{a} \times \vec{b}| \]

### Step 2: Analyze the left-hand side (LHS)
The left-hand side can be expressed as:
\[ |\vec{r} - \vec{a}| \cdot |\vec{b} - \vec{a}| \cdot \vec{n} \]
Since \(\vec{n}\) is a unit vector perpendicular to the plane formed by \(\vec{a}, \vec{b}, \vec{c}\), we can rewrite this as:
\[ |\vec{r} - \vec{a}| \cdot |\vec{b} - \vec{a}| \]
Thus, we have:
\[ |\vec{n} \cdot ((\vec{r} - \vec{a}) \times (\vec{b} - \vec{a}))| \]

### Step 3: Analyze the right-hand side (RHS)
The right-hand side is:
\[ \lambda |\vec{r} \times (\vec{b} - \vec{a}) - \vec{a} \times \vec{b}| \]
This represents the magnitude of the vector formed by the cross product of \(\vec{r}\) with \((\vec{b} - \vec{a})\) minus the cross product of \(\vec{a}\) and \(\vec{b}\).

### Step 4: Equate LHS and RHS
Equating both sides, we have:
\[ |\vec{n} \cdot ((\vec{r} - \vec{a}) \times (\vec{b} - \vec{a}))| = \lambda |\vec{r} \times (\vec{b} - \vec{a}) - \vec{a} \times \vec{b}| \]

### Step 5: Compare coefficients
By comparing the coefficients of both sides, we can deduce that:
\[ \lambda = |\vec{n}| \]
Since \(\vec{n}\) is a unit vector, we have:
\[ |\vec{n}| = 1 \]
Thus, we find:
\[ \lambda = 1 \]

### Conclusion
The value of \(\lambda\) is:
\[ \lambda = 1 \]